Calculating Volume from Revolved Area

[eberg]

I remember calculating volume by revolving shapes around axis using integration in school. I have an arbitrary shape shape that has no easily attainable equation, if I know the area of that shape, can I calculate the volume?

This is an example. I know this area, and want the volume if revolved around the y-axis:

http://img149.imageshack.us/img149/1/shapepo1.th.png

-Eric

 

[tiny-tim]

I have an arbitrary shape shape that has no easily attainable equation, if I know the area of that shape, can I calculate the volume?

Hi Eric!

No … compare the volume of a sphere with the volume of a cylinder with the same cross-section area.

Mathematically, volume = π∫(f(z))²dz, and area = ∫f(z)dz, and the integral of (f(z))² isn't a fixed multiple of the integral of f(z).

hmm … how about the surface area of the solid?

 

[eberg]

OK, thanks for the response.

 

[HallsofIvy]

There is a theorem (I can't quite remember the name right now) that the volume of a solid of rotation is equal to the area of region rotated and the circumference of the circle swept out by the centroid of the region. Of course, without knowing the formula for the boundary of the region you can't find that circumference exactly but for reasonably shaped regions you might be able to approximate it by seeing where lines from opposite points intersect.

 

[tiny-tim]

There is a theorem (I can't quite remember the name right now) that the volume of a solid of rotation is equal to the area of region rotated and the circumference of the circle swept out by the centroid of the region. Of course, without knowing the formula for the boundary of the region you can't find that circumference exactly but for reasonably shaped regions you might be able to approximate it by seeing where lines from opposite points intersect.

Hi HallsofIvy!

(i think you omitted the word "product" )

yes, the centroid would be at a distance of ((1/2)∫(f(z))²dz)/area,

so the volume is 2π times that times the area.

 

[HallsofIvy]

Thanks, tiny-tim.
And I now recall that it is "Pappus' Theorem".

 

[tiny-tim]

good ol' Pappus of Alexandria …

Thanks, tiny-tim.
And I now recall that it is "Pappus' Theorem".

Hi HallsofIvy!

At first, I thought, what's he talking about … isn't Pappus' theorem that projective hexagon thing?

Then I looked in http://en.wikipedia.org/wiki/Pappus'_Theorem#Theorems", and found that good ol' Pappus had two theorems …

Although Pappus's Theorem usually refers to Pappus's hexagon theorem, it may also refer to Pappus's centroid theorem.

so you're absolutely right!

 

[HallsofIvy]

Remarkable!

 

[tiny-tim]

No you can't use that word … it's copyright … see http://www.remarkable.co.uk/